Question

The diameter of a sphere is measured as $1.71cm$ using an instrument with a least count of $0.01cm$. The percentage error in surface area is?

Answer 1

Error in measurement refers to the deviation caused in the measurement from the actual measurement due to various reasons. Least count of a measuring instrument refers to the minimum value, which can be measured by that instrument. Least count of an instrument can also be defined as the error in measurement of that instrument. Percentage error refers to the ratio of least count of a measuring instrument to the measured value of physical quantity, multiplied by hundred.

Complete step by step answer:
When we use a measuring instrument to measure a particular physical quantity, it is quite possible that errors can occur in the measurements taken. These errors can be due to atmospheric conditions as well as personal reasons. Least count of a measuring instrument is the minimum possible value of physical quantity, which can be measured by that instrument. Least count of an instrument is also the least possible error, which can occur while taking measurements using that instrument. Mathematically, a measurement is usually represented as
$x=x'\pm error\Rightarrow x=x'\pm LC$
where
$x$ is the actual value of a physical quantity
$x'$ is the measured value of that physical quantity
$error=LC$, is the least possible error, which is nothing but the least count of the measuring instrument, with which the physical quantity is measured
Let this be expression 1.
Percentage error is an expression of error, in which the error is expressed in percentage. It is defined as the ratio of least count of a measuring instrument to the measured value of physical quantity, multiplied by hundred. Mathematically, percentage error is expressed as
$PE=$ $\dfrac{LC}{x'}\times 100$%
where
$PE$ is the percentage error caused while measuring a physical quantity using a measuring instrument
$LC$ is the least count of measuring instrument
$x'$ is the measured value of physical quantity, measured by the instrument
Let this be equation 2.
Coming to our question, we are given with the measured value of diameter of a sphere to be $1.71cm$, which is measured using an instrument whose least count is equal to $0.01cm$. We are required to determine the percentage error in surface area of the sphere.
We know that surface area of a sphere is given by
$S=4\pi {{r}^{2}}=4\pi {{\left( \dfrac{d}{2} \right)}^{2}}=\pi {{d}^{2}}$
where
$S$ is the surface area of a sphere
$r$ is the radius of the sphere
$d=2r$ is the diameter of the sphere
Let this be equation 3.
In equation 3, it is to be noted that diameter has to be recorded twice using the measuring instrument, to determine the surface area of sphere.
Let this be statement M.
Now, if $PE$ represents the percentage error in the surface area of the measured surface area of the given sphere, then, using equation 2, $PE$ is given by
$PE=2\left( \dfrac{LC}{d'} \right)\times 100$%
where
$PE$ is the percentage error caused while measuring the surface area of sphere using the given measuring instrument
$2$ is the multiplication factor, according to statement M
$d'$ is the measured diameter of the sphere
$LC$ is the least count of the instrument used, using equation 2
Let this be equation 4.

Substituting the given values from the question in equation 4, we have
$PE=2\left( \dfrac{LC}{d'} \right)\times 100$%$\Rightarrow PE=2\times \dfrac{0.01cm}{1.71cm}\times 100$%$=1.169$%
Let this be equation 5.

Therefore, from equation 5, we can conclude that the percentage error in surface area of the given sphere, when measured using the given instrument is equal to $1.169$%.

Note: Students need not get confused about the multiplication by two, done in equation 4. Here, change in surface area is determined with the help of change in diameter of the sphere, measured by the given instrument. From equation 3, it is clear that diameter has to be measured twice, to determine the surface area of the sphere. It is these two measurements, which comes in the form of a multiplication factor of two, in equation 4, while determining the percentage error in surface area of the sphere.